The equations are like this: 1-X=exp(-A*Y^B)
I need to derive the parameters A and B and use the same model to predict the values, which should be compared with the experimental observations.
The solved equation is: Ln (X) = {Ln(-Ln(1-X))-Ln (A) }/ B
Linear regression applied to transformed equation only looks easy. In fact, any nonlinear transformation introduces following undesirable effects:
1. when the original data are not correlated then their transformed version may not enjoy this property,
2. experimental uncertainties are usually symmetric around measured values; after non-linear(!) transformation this property is lost.
The lack of correlations and symmetric (Gaussian) distribution of of uncertainties are the main assumptions on which the linear regression is based.
The two features of nonlinear transformations, shown above, conspire to deform the desired result. In some cases also rounding errors are the source of severe distortion of the results obtained this way (think of functions like tanh(x) or 1/x computed for large |x|).
Conclusion: linear regression on transformed data often produces only very crude approximations to true values. Nevertheless, such results are useful as a start point for more advanced (nonlinear) fitting routine, but should never be regarded as final ("best fit").
The phenomenon you observed was extensively researched for the data fitted to power law (y ~ a*x^b), often encountered as the tails of various distributions Zipf's law, for example). Here, only the parameter 'b' is interesting but its value appears very unreliable after 'linearized fit'.
Sir, Thank you for your comments. It is a mistake from my side and u r right. It is Ln(Y) on the LHS. I have tried it in Excel with logarithmic regression analysis (graphical method). The values i have got does not match the experimental values. I tried with some known answers, I have tried it again. Is it possible for you to explain this.
Actually the equation was like this: 1-X=exp(-A*Y^B)
I got a solution from sources like this Ln (Y) = {Ln(-Ln(1-X))-Ln (A) }/ B
I made plot of Ln(Y) against Ln(-Ln(1-X)), which should give a slope and intercept. I found out Band A from slope and intercept respectively. But when i am applying back the values to the equation, I am not getting values equivalent to the experimental values. This is the problem.
Linearizing this equation is perhaps not a good way to do a curve fit for an equation like this. The problem is when you have data that is near the asymptote (Y is large), taking the logarithm will amplify the noise in that data. In the end, you have performed an extreme weighting.
My opinion is that this has to be treated as a nonlinear regression.
Linear regression applied to transformed equation only looks easy. In fact, any nonlinear transformation introduces following undesirable effects:
1. when the original data are not correlated then their transformed version may not enjoy this property,
2. experimental uncertainties are usually symmetric around measured values; after non-linear(!) transformation this property is lost.
The lack of correlations and symmetric (Gaussian) distribution of of uncertainties are the main assumptions on which the linear regression is based.
The two features of nonlinear transformations, shown above, conspire to deform the desired result. In some cases also rounding errors are the source of severe distortion of the results obtained this way (think of functions like tanh(x) or 1/x computed for large |x|).
Conclusion: linear regression on transformed data often produces only very crude approximations to true values. Nevertheless, such results are useful as a start point for more advanced (nonlinear) fitting routine, but should never be regarded as final ("best fit").
The phenomenon you observed was extensively researched for the data fitted to power law (y ~ a*x^b), often encountered as the tails of various distributions Zipf's law, for example). Here, only the parameter 'b' is interesting but its value appears very unreliable after 'linearized fit'.
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